# Chapter 2 Properties of pure substances

## 2.1 Properties

Substances have properties that are used to identify and describe them. You should be familiar with four properties from everyday life: temperature, pressure, volume and mass. We measure the amount of milk by volume and meat by mass. We quantify the “hotness” or “coldness” of air by measuring its temperature. Pressure gives an indication of the force necessary to contain a fluid.

We distinguish between intensive and extensive properties. Temperature is an intensive property as its value does not depend on the amount of matter of the substance. A cup of water can be at 20\(^\circ\)C as well as a drop of water. Volume and mass are extensive properties. The mass and volume of a substance is directly proportional to the amount of matter making up the substance. A cup of water cannot have the same mass as a drop of water.

Another characteristic of properties is that the value of a property at the present is not dependent on the history of the substance. The temperature of a cup of water at the present moment is not dependent on its temperature a few minutes ago. The water could have been colder or hotter at a previous stage. Therefore, properties are called point functions. Distance on the other hand, is a path function as the distance traveled to get from point A to point B, is dependent on the path followed. Work performed and heat transferred are also path functions as will become clear later.

### 2.1.1 Temperature

Temperature is measured in degrees Celsius [\(^\circ\)C] or Kelvin [K].
To convert a temperature in \(^\circ\)C to K, add 273.15.
This means 50\(^\circ\)C is the same as 323.15K.
It also means that a temperature difference expressed in degrees Celsius is numerically the same as the difference expressed in Kelvin.
The lowest possible temperature is 0K.
Temperature is an intensive variable.^{1}

### 2.1.2 Pressure

Pressure (\(P\)) is the force per unit area: \[P = \frac{F}{A}\]

With F in [\(kN\)] and A in [\(m^2\)], P will be in [\(\frac{kN}{m^2}\)]=[\(kPa\)]. Pressure is an intensive variable.

Other common pressure units are:

atmosphere (\(1 \mbox{atm} = 101.325 kPa\)) and

bar (\(1 \mbox{bar} = 100 kPa\))

#### 2.1.2.1 Absolute or total pressure

The total pressure is the total force exerted on a surface divided by the area of the surface. In a gas, the force exerted on a surface is due to the collision of the gas molecules against the surface. There is a limit at how low total pressure can be. If no force is exerted on a surface the total pressure is zero. The pressure of a complete vacuum is zero because there are no molecules colliding with the surface of the container. Absolute pressure (\(P_{abs}\)) is the pressure measured above this zero point. The value of total pressure can never be negative.

#### 2.1.2.2 Atmospheric pressure

Atmospheric pressure (\(P_{{atm}}\)) is the total pressure caused by the weight of the atmospheric air.
This pressure varies according to location and weather patterns.
The average value at sea-level is \(101.325kPa\).
The average atmospheric pressure for Potchefstroom is \(87.0kPa\) and \(99\%\) of the time the pressure will be between \(86kPa\)
and \(88kPa\).
The average atmospheric pressures for a few major cities in South Africa are given in Table 2.1.^{2}
It is clear that the height above sea level plays a very important role in determining the average value.^{3}
Atmospheric pressure is also called ambient pressure.

City | Height above sea level [\(m\)] | Mean atmospheric pressure [\(kPa\)] |
---|---|---|

Kimberley | 1230 | 88.5 |

Polokwane | 1262 | 88.3 |

Pretoria | 1350 | 87.5 |

Potchefstroom | 1351 | 87.0 |

Bloemfontein | 1395 | 86.7 |

Johannesburg | 1700 | 83.5 |

#### 2.1.2.3 Gauge pressure

Gauge pressure (\(P_g\)) is the difference between the total pressure inside the vessel and an the total pressure outside. \[P_{gauge}=P_{inside}-P_{{outside}}\] Normally the total pressure outside is equal to atmospheric pressure but for a vessel submerged in water such as a submarine, or the air cylinder of a Scuba diver, the total pressure outside may differ from the atmospheric pressure. If the pressure inside is lower than the pressure outside, the gauge pressure will be negative and when it is higher, it will be positive. Except when specifically stated that it is a total or absolute pressure gauge, the pressure displayed by a pressure gauge is the gauge pressure. We use a pressure gauge at the petrol station to measure the gauge pressure of the air in the tires of our cars.

**Example**

Assume you have inflated the tyres of your car to a gauge pressure of \(200kPa\) on the Highveld where the atmospheric pressure on that day is \(85kPa\). The absolute (or total pressure) inside the tyres is now \(285kPa\). You now drive to the sea where the atmospheric pressure on that day is \(101kPa\). What is the gauge pressure of the tyres now?

*Solution*

If the tyre does not leak, the mass of air inside does not change. Let us assume the volume stayed the same as well as the temperature. Therefore the absolute pressure inside the tyres also remained the same (this can be shown to be true by using the Ideal Gas Law which is done later), but the gauge pressure would have dropped to 184kPa. In order to suspend the weight of your car at this lower pressure, a bigger area is required and your tyres will appear a little flatter than at the highveld. You will have to inflate the tyres.

#### 2.1.2.4 Hydrostatic Pressure

The pressure exerted by a liquid column with a uniform density, is given by the equation:

\[P=\frac{\rho g h}{1000}\]

With \(\rho\) in [\(\frac{kg}{m^3}\)], \(g=9.81\frac{m}{s^2}\) and h in [\(m\)], P will be in [\(kPa\)].

**Example**
Assume you go down a mine shaft \(1000m\) deep.
Calculate the pressure exerted by a column of air \(1000m\) high.
Assume the air has a constant density of \(1\frac{kg}{m^3}\).

*Solution*
\[P=\frac{1\times9.81\times 1000}{1000}=9.81kPa\]
This is in the same range as the change in pressure you will experience traveling from the Highveld to the coast.
The density of the air will change due to the significant change in pressure - so the assumption of uniform air density can be questioned.
How will you account for the change in air density?

**Example**

A sectional view of a U-shaped manometer filled with mercury connected to a pipe filled with air is shown in Figure 2.1. The difference in height of the mercury in the legs of the manometer (h) is \(0.2m\). The density of the mercury can be taken as \(13600\frac{kg}{m^3}\) and the atmospheric pressure as \(87kPa\). Calculate the absolute and gauge pressure inside the pipe.

*Solution*

The pressure exerted by a column of mercury with height \(h\) is: \[P_{mercury}=\frac{0.2 \times 9.81 \times 13600}{1000}= 26.68kPa\] The pressure on the surface of the mercury exposed to the atmosphere is equal to \(87kPa\). In the other leg of the manometer, at the same elevation, the pressure will also be \(87kPa\). Moving upwards from that elevation to the surface of the mercury, will cause a drop in total pressure. Therefore:

\[P_{abs,inside}=87-26.68=58.32kPa\]

The density of the air is very low, therefore it can be assumed that the pressure inside the pipe is uniform. The gauge pressure is \(-26.68kPa\). From the relative height of the mercury columns in the legs of the manometer it is also clear that the pressure inside the pipe is lower than the pressure inside, therefore it makes sense that the gauge pressure has a negative value.

### 2.1.3 Volume and Mass

It is often convenient to combine the two extensive properties, volume and mass, to form an intensive property: specific volume (\(v\)). Specific volume is the total volume divided by the total mass \(v=\frac{V}{m}\) [\(\frac{m^3}{kg}\)]. Specific volume is an intensive property. Density (\(\rho\)) is the inverse of specific volume.

**Example**

A cup holds \(250m\ell\) of boiling hot water at \(95^\circ C\). The mass of the water is \(240.4g\). Calculate the specific volume in [\(m^3/kg\)] and the density in [\(kg/m^3\)].

*Solution*
Specific volume is the total volume divided by the total mass: \(v=V/m\).
First determine the values of \(V\) and \(m\).

\[\begin{aligned} V &= 250m\ell \times \frac{1\ell}{1000m\ell} \times \frac{1m^3}{1000\ell} \\ &= 250 \times 10^{-6} m^3 \\ m &=240.4g \times \frac{1kg}{1000g}\\ &= 240.4 \times 10^{-3} kg\\ \therefore v &=\frac{250\times 10^{-6}}{240.4 \times 10^{-3}}\\ &= 0.00104 \frac{m^3}{kg} \\ \mbox{and} \, \rho &=\frac{1}{v} = 961.5 \frac{kg}{m^3} \end{aligned}\]

Sometimes the density of a substance is given relative to a reference substance. Then it is called relative density or the somewhat misleading term, specific gravity (SG). In the case of liquids (and often solids) the reference substance is usually water at a pressure of \(101.3kPa\) and \(4^\circ C\) with a density of \(1000 \frac{kg}{m^3}\). The SG of mercury is usually taken as 13.6. This means its density is \(13600\frac{kg}{m^3}\)

Let us see how \(P,T\) and \(v\) are interdependent.

## 2.2 Single phase systems

The term ** phase** refers to a quantity of matter that is homogeneous throughout in both chemical composition and physical structure.
Homogeneity in physical structure means that the matter is all solid, or all liquid, or all gas.
An oil-water mixture consists of two liquid phases - a water phase and an oil phase.
They are both liquids but their chemical compositions differ.
The three phases of water are shown in Figure 2.2 below.
The three phases are separated by phase boundaries with zero thickness - although to be able to see it on the graph, a line with finite thickness must be used, otherwise the line would have be invisible.

^{4}It means that when you specify the value of temperature

**and**the value of pressure, that the point can never fall on the line. For instance for \(P=87.00kPa\) and \(T=95.78^\circ C\) water will be a liquid and at the same pressure but a slightly higher temperature of \(T=95.79^\circ C\) it will be a gas. You can use as many digits as you want to, the water will be either a liquid or a gas. (Especially doing calculations using software, it is easy to forget this.) It will become clear later that the lines in Figure 2.2 represents two-phase mixtures, either liquid/gas, liquid/solid or solid/gas.

For single-phase substances, we can alter the values of two intensive properties independently - in other words, a change in temperature for instance, will not necessarily lead to a change in pressure (as long as we stay clear of the phase boundaries!) For instance, consider liquid water at \(100kPa\) and \(50^\circ C\) (indicated by the \(\Box\)). When we heat a cup of cold water in the microwave oven, we increase the temperature while the pressure exerted by the atmosphere on the water remains constant. When we put the cup in a refrigerator, the temperature will now drop but the pressure will not change. We can also vary the temperature and pressure of steam (shown by the \(\circ\)) independently of each other.

Because we can change the value of two intensive properties independently we say a single-phase substance has two degrees of freedom.^{5}
It means that we only have to specify the values of two independent intensive properties in order to fix the state of the system and the values of the other intensive properties.
The ** state** of the substance is simply the condition of the substance as described by its properties.
If we heat up water, we say its state has changed because its temperature has changed.
There are different ways in which to determine the value of the other properties.

### 2.2.1 Ideal gases

An ideal gas is a hypothetical gas whose molecules/atoms do not attract or repulse each other and collide elastically with each other and the
walls of the container.
They are point particles occupying negligible space.
At high temperatures and low pressures,^{6} the density of gases are low (the particles will occupy negligible space) and the kinetic energy of the particles much higher than any possible inter-particle interactions.
Under these conditions gases approach ideal gas behavior.
For an Ideal Gas, the relationship between the pressure, temperature and total volume are given by the Ideal Gas Law: \[PV=n\overline{R}T\] where
\(\overline{R}\) is the universal gas constant with a value of \(8.3145 \frac{kPa\cdot m^3}{kmol \cdot K}\).
It is important to note that the molecular mass of the gas under consideration does not play a role in this equation.
A kilomole of Helium and and a kilomole of Air at the same pressure and temperature will occupy the same volume - at \(25^\circ C\) and \(100kPa\) that volume will be \(24.8m^3\) although the Helium will weigh 4kg and the Air 28.97kg.
This means a balloon filled* with one kmole of Helium will be able to lift a mass of almost \(25kg\).
The diameter of such a spherical balloon will be \(3.4m\).^{7}

In Engineering we prefer working with mass. The Ideal Gas Law is rewritten as \(PV=mRT\) with \(R=\frac{\overline{R}}{MM}\). MM is the molecular mass, and \(R\) is the ideal gas constant for the substance under consideration. Its units are \([kPa\cdot m^3\cdot kg^{-1}\cdot K^{-1}]\) and it has a specific value for every substance. The ideal gas constant for three substances are shown in the Table 2.2 below.

Substance | R |
---|---|

Air | 0.2870 |

Steam | 0.4615 |

Helium | 2.0771 |

Tables with the values of the ideal gas constant for different substances can be found in literature (Table A.5).^{8}(Sonntag and Borgnakke 2012).
It is often convenient to work with specific volume and the Ideal Gas Law then becomes: \[Pv=RT\]

An ideal gas is a single phase substance and according to the Gibbs phase rule, has two degrees of freedom. This is also clear from the Ideal Gas Law that once we know what the values of the temperature and pressure are, we can calculate the specific volume. If fact if we know the values of any two variables (\(P,T\); \(P,v\); \(T,v\)) we say the state is fixed and we can calculate the value of all other intensive properties.

**Example**

Consider air at \(25^\circ C\) and \(101kPa\).
Calculate its specific volume and the mass of \(1m^3\) of air.

*Solution*

\[\begin{aligned}
v &= \frac{RT}{P} \\
&= \frac{0.287\times298.15}{101} \\
&= 0.8475 \frac{m^3}{kg}\\ \\
m &=\frac{V}{v}\\
&=\frac{1}{0.8475}\\
&=1.180 kg\\
\end{aligned}\]

Make sure of the units when using the Ideal Gas Law.
Temperature is in Kelvin, pressure is the absolute pressure in \(kPa\) and specific volume is in \(m^3/kg\).
In the case of gases like **air** in everyday use and **Helium**, the Ideal Gas Law is quite accurate.

In this notes, Helium and Air are always treated as ideal gases.

### 2.2.2 Real Gases

Super-heated steam is usually considered to be a real gas^{9} and the specific volumes at different temperatures and pressures are tabulated in tables known as the steam tables.
These are available in hard copy or as software.

**Example**

Find the tables for super-heated steam at the back of your textbook.(Sonntag and Borgnakke 2012) (Note that water’ refers to the chemical \(H_2O\) and is not an indication of the phase.

We talk of solid water, liquid water and water vapor when we refer to the solid, liquid or gas phases.)

Determine the specific volume of steam at \(5000kPa\) and \(350^\circ C\). (\(0.05194 m^3/kg\)).

Now use the Ideal Gas Law to determine the specific volume (\(0.0575 m^3/kg\)). You can see there is a significant difference between the two values.

**Example**

Steam at \(5000kPa\) has a specific volume of \(0.054 m^3/kg\).
Find the temperature.

*Solution*

You will note that in the table for super-heated water at \(5000kPa\), \(0.054 m^3/kg\) lies between the specific volumes at \(350^\circ C\) and \(400^\circ C\).
You need to interpolate to find the required temperature.
(\(367.5^\circ C\))

### 2.2.3 Liquids and solids

The specific volume of compressed liquid water as function of temperature and pressure can be found in the steam tables.
See Borgnakke Table B.1.4 (Sonntag and Borgnakke 2012).
Note that an increase in pressure has almost no effect on the specific volume of liquid water.
For a ten fold increase in pressure (from \(500kPa\) to \(5000kPa\)) for water at \(20^\circ C\) the specific volume changes from \(0.001002 m^3/kg\) to \(0.001000 m^3/kg\).
Therefore, liquids (and solids) are usually considered to be in-compressible.
From the table it is also clear that the specific volume of water does increase somewhat with an increase in temperature.
It is therefore common practice to take the specific volume of compressed liquid water at \(T\) and \(P\), to be equal to the specific volume of *saturated* liquid water at temperature \(T\), even though the pressure of saturated liquid water will be different to \(P\).

The volume of saturated solid water (ice) as function of temperature can also be found in Borgnakke Table B.1.5 (Sonntag and Borgnakke 2012). Ice can also be assumed to be in-compressible. For other substances, detailed tables are usually not available. However, the density of several liquids and solids at room temperature can be found in textbooks Table A.3 A.4 (Sonntag and Borgnakke 2012).

**Example**

Find the specific volume of water at \(120^\circ C\) and \(5000kPa\) using the saturated water tables.

*Solution*

Using the temperature and pressure of the water to find the position of the water on a pressure-temperature diagram (Figure 2.2), it is clear that the water is a compressed liquid.
The specific volume of saturated liquid water at \(120^\circ C\) is \(0.001060m^3/kg\).
Assuming liquid water to be in-compressible, the specific volume at \(5000kPa\) is therefore also taken as \(0.001060m^3/kg\) even though the saturation pressure is only \(270.1kPa\).

It is left to the reader to compare this value to the value at \(5000kPa\) and \(120^\circ C\) from the Compressed liquid Table B.1.4 (Sonntag and Borgnakke 2012).

## 2.3 Two-phase systems

We will first consider a system consisting of liquid water and water vapor.

### 2.3.1 Phase change

If we heat up liquid water enough it will start to boil - it will undergo a phase change. It is instructive to study some of the events that occur as a pure substance undergoes a phase change. Consider water at \(101.325kPa\) in a piston-cylinder assembly. The water is at 20\(^\circ\)C, a temperature lower than the boiling point and therefore it is a single-phase system. It has two degrees of freedom. The temperature and the pressure can be varied independently without the water changing its phase. The water is heated while the pressure remains constant at 101.325\(kPa\). The volume increases slightly as the water is heated and the temperature rises. (Assume the combined effect of the cylinder and the ambient pressure is 101.325\(kPa\) and the piston is frictionless and can therefore move to accommodate any increase in volume.)

At the boiling point (\(100 ^\circ C\)), vapor will start to form. At the boiling point, when the first molecule of liquid water is about to be transformed into a vapor, the quality (see Paragraph 1.3.2) is still equal to zero and the liquid is called a saturated liquid. As more heat is added more vapor forms. All the heat that is added to the system is used to change the liquid into vapor and the temperature remains constant. It is no longer possible to vary temperature and pressure independently without the system changing phase. If we increase the pressure on the system (by for instance putting additional weights on the piston) the vapor will condense, release heat, and a single-phase system will result once more. If we resume heating at this higher pressure, the temperature will rise to a higher value (as previously) before vapor will start to form again. All the liquid will eventually vaporize. When the last liquid molecule vaporizes, the phase is called a saturated vapor. If the saturated vapor is heated, its temperature will rise and super-heated vapor will form. The constant pressure heating of liquid water at various pressures is shown as dashed lines in Figure 2.3 below.

If the pressure is high enough, no phase separation will take place. The liquid will change from a liquid-like phase to a gas-like phase without two phases forming. You can see this happening in Figure 2.3 at \(40000kPa\). The minimum pressure where no phase separation takes place, is called the critical pressure. The temperature at the critical pressure is called the critical temperature. The critical pressure of water is \(22.09MPa\) and the critical temperature \(374.14^\circ C\). As long as two phases are present, the system is at the saturated state. The liquid is called saturated liquid and the gas, saturated vapor. The temperature is the saturation temperature and the pressure, the saturation pressure or vapor pressure. This condition is often also called an vapor-liquid equilibrium mixture.

On a \(P-v\) diagram Figure 2.5 the constant pressure heating of water will be a horizontal line. The constant pressure heating of water can also be shown as a horizontal line on a T-P phase diagram (Figure 2.2) The liquid-gas phase boundary in Figure 2.2 ends in the critical point.

It is clear from Figure 2.2 that the equilibrium pressure is a unique
function of temperature^{10} and that, for a two-phase mixture,
temperature and pressure are no longer independent. Once the pressure
(or temperature) is specified, the temperature (or pressure) and the
specific volumes of the saturated vapor and the saturated liquid are
fixed. This seems to imply that vapor/liquid two-phase system only has
one degree of freedom because all the variables are fixed once one
variable is fixed. This is true for the intensive variables:
\(T_{sat}, P_{sat},\) \(v_{sat,vapor},\) \(v_{sat,liquid}\).

**Example**

Find the saturation pressure, specific volume of saturated water vapor
and specific volume of saturated liquid water at \(100^\circ C\). The
saturation pressure as function of temperature can be found in Table
B.1.1 of Cangel (Cangel and Boles 2002) (\(101.325kPa; 1.6729 m^3/kg; 0.001044 m^3/kg\))

In general we need to know the relative amount of liquid and vapor. We therefore need another variable - bringing to two the number of variables necessary to fully quantify the system. The second variable can be specific volume of the mixture or a variable that we will call quality.

### 2.3.2 Quality

For a saturated liquid-vapor mixture, the quality \((x)\) is the ratio of
the mass of vapor to the total mass of the two-phase mixture:^{11}

\[x=\frac{m_{vapor}}{m_{liquid}+m_{vapor}}\]

The value of the quality varies from zero to unity. For a saturated
liquid, \(x=0\), and for a saturated vapor, \(x=1\). Quality is only defined
for the saturated condition.^{12} Using the saturation tables and the
definition of quality, the specific volume of a two-phase liquid-vapor
mixture can be determined:

\[\begin{equation} v=(1-x)v_f +x v_g \tag{2.1} \end{equation}\]

**Example**

Determine the pressure and specific volume of water at 100\(^{\circ}\)C
with a quality of \(0.7\).

*Solution*

Because the quality has a value between zero and one, we know it is a
two-phase mixture and the pressure will be the saturated pressure at
100\(^{\circ}C\), \(101.3kPa\). The specific volume of the two phase mixture
is calculated using Equation 1.1 and the volumes of saturated liquid and
saturated vapor at 100\(^\circ C\). ( \(1.171343 m^3/kg\))

It is also possible to get liquid/solid equilibrium and vapor/solid equilibrium – as shown in Figure 2.2. It is also possible to get equilibrium between three phases - vapor, liquid and gas. It is the triple point and according to the Gibbs phase rule, a pure substance has only one such triple point: at the intersection of the three phase boundary lines – as can be seen in Figure 2.2.

If the pressure, temperature and specific volume characteristics of
water are plotted on three perpendicular axes, one each for temperature,
pressure and specific volume, a three dimensional shape results as shown
in Figure 2.4 below.^{13}

## 2.4 Determining the phase of water

It is necessary to provide numerical values for two of a possible four variables in order to fix the state of a substance. The four variables are: Pressure, Temperature, Specific volume and quality. In order to determine the values of the unknown variables, first we first need to determine the phase (solid, liquid or gas) of the substance. Once we know the phase, we know which table (compressed liquid, saturated liquid, super-heated vapor) to use. Let us look at common combinations of variables used to fix the state and how to determine the phase.

### 2.4.1 Temperature and pressure

The numerical values for temperature and pressure are supplied/specified. The phase is determined by comparing the specified temperature to the saturation temperature at the specified pressure and/or comparing the specified pressure to the saturation pressure at the specified temperature. In effect this means plotting the state on a Pressure–Temperature Phase diagram. Only three phases are possible when the pressure and temperature are specified: solid, liquid and gas. As mentioned before, the lines separating the phases as in Figure 2.2 has zero thickness and the state will either be a solid OR a liquid OR a vapor.

The slope of the solid/liquid equilibrium line for water on a
pressure-temperature graph has a very large negative slope. This means
liquid water under high pressure will freeze at a slightly lower
temperature than water under a lower pressure. (Most other substances
contracts upon freezing and the slope of the solid/liquid boundary is
positive.)^{14} Due to the negative slope of the solid/liquid phase
boundary, the temperature of the triple point of water is \(0.01^\circ C\)
while the melting point of ice at \(100kPa\), is \(0^\circ C\). In this
notes it is assumed that the phase boundary between ice and liquid water
is vertical and that ice melts at \(0^\circ C\) at any pressure.^{15}

Consider water at \(100kPa\) and \(50^\circ C\) – the \(\Box\) in Figure 2.2. The
specified temperature is lower than \(99.62^\circ C\), the saturation
temperature at \(100kPa\)^{16} and the specified pressure higher than
\(12.34kPa\), the saturation pressure at the specified temperature. The
phase is therefore a compressed (or sub-cooled) liquid as is also
evident from its position on the phase diagram. The quality is undefined
and if we assume an in-compressible liquid, the volume of the substance
is the same as that of the saturated liquid at \(50^\circ C\),
\(0.001030\frac{m^3}{kg}\), which can be read from the tables for
saturated water (Table B.1.2).(Sonntag and Borgnakke 2012)

Consider water at \(100kPa\) and \(150^\circ C\) – the \(\circ\) in Figure 2.2. Its pressure is lower than \(475.9kPa\), the saturation pressure at the specified temperature and its temperature is higher than \(99.62^\circ C\), the saturation temperature at the specified pressure. This means it is an super-heater vapor (often also called a gas), which is also evident from its position on the phase diagram. Again the quality is undefined and the specific volume can be read from the super-heated water vapor tables, \(1.93636\frac{m^3}{kg}\)

### 2.4.2 Pressure or temperature and specific volume

The heating of compressed liquid water at a constant pressure (lower
than the critical pressure) will first result in a saturated liquid,
then in a two-phase mixture, then in a saturated vapor and then in a
super-heated vapor. The P-\(v\) phase diagram of water is shown in Figure
2.5 below.
The solid water phase (ice) has been omitted from this phase diagram.
Because liquid water expands upon freezing (as can be seen in Figure
2.4), the P-v
phase diagram showing the solid phase, would have been unnecessarily
complicated.^{17}

Consider water at \(5000kPa\) and \(0.001000\frac{m^3}{kg}\) – the
\(\Box\) in Figure 2.5. It is a compressed liquid because its specific
volume is smaller than the specific volume of the saturated liquid at
the same pressure.^{18} For a compressed liquid, quality is undefined.
As the specific volume of a compressed liquid does not change much with
changes in temperature at lower values of temperature – as can be seen
in Figure 2.3, it may be difficult to obtain an accurate
value for temperature from specified values of pressure and specific
volume.

Water at \(100kPa\) and \(0.01\frac{m^3}{kg}\) (the \(\diamondsuit\) in Figure 2.5) is a two-phase mixture as its specific volume lies between that of the saturated liquid and saturated vapor at \(100kPa\). Its quality can be calculated from Equation (2.1) using the specific volume of the saturated liquid (\(0.001043\frac{m^3}{kg}\)), and the specific volume of the saturated vapor at \(100kPa\) (\(1.694\frac{m^3}{kg}\)) to obtain a value of \(x=0.005291\). The temperature is the saturation temperature at \(100kPa\), \(99.62^\circ C\).

Water at \(100kPa\) and \(4.02781\frac{m^3}{kg}\) (the \(\circ\) in Figure 2.5) is a super-heated vapor as its specific volume is larger than that of a saturated vapor at the specified pressure. Its temperature can be determined from the super-heated water tables. In this case the temperature can be read directly from the tables (\(600^\circ C\)), but in general, interpolation will be necessary. Quality is undefined for a super-heated vapor.

When the temperature and specific volume are specified, the phase must first be determined by comparing the value of specified volume to the specific volume of the saturated liquid and and the specific volume of the saturated vapor at the specified temperature - in effect by plotting the state on a \(T-v\) diagram (Figure 2.5).

### 2.4.3 Temperature or pressure and quality

The quality of a saturated liquid is \(0\) and for a saturated vapor, \(1\). As quality is undefined for a sub-cooled liquid and super-heated vapor, if the quality is specified, it means it is a two-phase mixture and the temperature (or pressure) for the specified pressure (or temperature) can be read from the tables for saturated water and the specific volume determined from Equation (2.1). An example is done in Paragraph 2.3.2

## 2.5 Unit conversion

It has been stated previously that a pressure of one atmosphere is the same as \(101.325kPa\). Let us assume we know the pressure is \(100 atm\) and now need to know what the pressure is in \(kPa\). Because \(1atm=101.325kPa\), we know \(\frac{101.325kPa}{1atm}=1\). Now:

\[100atm = 100 atm \times \frac{101.325 kPa}{1 atm}= 10100 kPa\]

Note that the answer can only contain three significant digits. Sometimes psi (pound-force per square inch) is used as a pressure unit. Let us calculate the value (in kPa) for a pressure of 14.50psi. A pound-force is the weight of a mass of one pound in the gravitational field of the earth.

\[14.50psi=\frac{14.50lb_f}{1(inch)^2} \times \frac{4.448N}{1 lb_f} \times \frac{1( inch)^2}{(0.02540m)^2}=99.970kPa\]

Note that as only four significant digits is used the calculation, the answer can also not be more that four significant digits.

You can ask Google: ‘Convert psi to kPa’ and get the conversion factor directly.

Temperature can also be measured in \(^\circ F\). \(T[^\circ F]=1.8T[^\circ C]+32\). This means that \(100^\circ C= 212^\circ F\). The absolute temperature associated with \(^\circ F\) is Rankine. The temperature in [R] is obtained by adding 459.67 to the temperature in [\(^\circ F\)]↩︎

With acknowledgment to the South African Weather Service.↩︎

Early explorers used the boiling point of water to determine their height above sea level. The boiling point of water at \(87kPa\) is \(95.78^\circ C\), which corresponds (for Potchefstroom) to a height above sea level of 1351m. This means the boiling point of water changes on average approximately by \(0.3^\circ C / 100m\) between sea level and this height. It is clear from Figure 2.2 that the boiling point of water does not change linearly with pressure but rather according to the Antoine equation which is discussed in a footnote of Paragraph 1.3.1.↩︎

Make sure that you realize and remember the implication of the fact that the line has zero thickness.↩︎

The number of degrees of freedom are determined by the Gibbs Phase Rule. The Gibbs Phase Rule states that the number of independent intensive variables is equal to 3 minus the the number of phases. See Cangel (p.759) (Cangel and Boles 2002). In the case of a single phase system, the degrees of freedom are therefore \(3-1=2\).↩︎

As a rule of thumb, a gas can be considered ideal if its temperature is twice its critical temperature and its pressure half its critical pressure. Critical pressure and temperature are explained in Paragraph 1.3.1.↩︎

The diameter of a typical hot-air balloon is \(16m\) and it can lift approximately \(600kg\), depending on the temperature difference between the air inside and the air outside the balloon. Check your understanding by making a few assumptions and see whether you can get the same figure.↩︎

In literature, the units of \(R\) are usually given as [\(\frac{kJ}{(kg\cdot K)}\)]. It is left to the reader to show that this is the same as [\(\frac{(kPa\cdot m^3)}{(kg\cdot K)}\)].↩︎

Optional textbook reading: Cangel (Cangel and Boles 2002) discuss the question: Is water vapor an ideal gas?’ and Compressibility factor - a measure of deviation from Ideal Gas Behaviour’. See also paragraphs 3.6 and 3.7 in Borgnakke (Sonntag and Borgnakke 2012) on the same topics. Even though you can find the ideal gas constant for steam and other real gases in the tables giving the ideal gas properties of gases, the Ideal Gas Law is only used for these substances under special conditions. For instance atmospheric air contains some water vapor and this water vapor is usually assumed to obey the Ideal Gas Law.

**In this notes, ideal gas behavior is never assumed for steam even though there may be times that is a good assumption.**↩︎The relationship between the equilibrium pressure \((P)\) and the equilibrium temperature \((T)\) (in Kelvin) of a two-phase system, is given by the Clausius-Clapeyron equation which when integrated, is of the form: \(ln P=A-\frac{B}{T}\). The values of A and B can be obtained from experimental values by determining the slope and intercept of a straight line obtained when plotting \(ln(P)\) against \(1/T\). The Antoine equation is an empiricized version of this relationship between equilibrium pressure and equilibrium temperature.↩︎

Quality cannot be used in the Gibbs Phase rule because quality is not an intensive variable. However, the Gibbs Phase rule can be applied to each phase separately. The relative amount of liquid and vapor are defined by the quality or specific volume of the mixture.↩︎

This means that \(x\) can never have a value bigger than unity such as \(1.2\). If you ever calculate \(x\) and find it is bigger than \(1\), you have made a mistake somewhere!↩︎

Make sure that you can see that the different phase diagrams (P-\(v\); T-\(v\) and P-T) are two-dimensional perspectives of this three-dimensional surface. You should be able to see that the phase boundary between the liquid and vapor/gas phases, which is a line on a \(P-T\) phase diagram, is a bell-shaped surface on a T-\(v\) and P-\(v\) diagram. You should also be able to see that the line (between the saturated liquid and the saturated vapor) on the T-\(v\) and P-\(v\) diagrams, is a point on the vapor-liquid phase boundary on the \(P-T\) diagram.↩︎

Phase diagrams are discussed in greater detail in Paragraph 2.7of Borgnakke (Sonntag and Borgnakke 2012).↩︎

The Clapeyron equation can be used to calculate the melting point of ice at pressures smaller than \(10 000kPa\). At a pressure of \(7000 kPa\), the melting point of ice will be \(-0.5^\circ C\).(www.chm.bris.ac.uk)↩︎

You can find the saturation values in steam tables in the back of your textbook.↩︎

Liquid water expands almost 10% upon freezing. This expansion can cause water pipes to burst when the liquid water inside freezes on cold winter nights. Consider liquid water in equilibrium with solid water (ice) The specific volume of liquid water (up to \(5000kPa\), rounded to four decimals figures), is \(0.0010 m^3/kg\) while the specific volume of solid water in equilibrium with liquid water, rounded to four decimal figures, is \(0.0011 m^3/kg\).↩︎

The specific volume of compressed liquid water changes almost nothing with pressure and very little with temperature. Have a look in the compressed liquid table in Borgnakke Table B.1.4 (Sonntag and Borgnakke 2012). In order to get from the \(\Box\) in Figure 2.5 (\(v=0.0010m^3/kg\)) to the saturated liquid line at \(5000kPa\) (\(v=0.001286m^3/kg\)), required a temperature increase from \(20^\circ C\) to \(264^\circ C\).↩︎